Problem: Let $a(x)=3x^4-2x^2+x+5$, and $b(x)=x^4+x^2+x+1$. When dividing $a$ by $b$, we can find the unique quotient polynomial $q$ and remainder polynomial $r$ that satisfy the following equation: $\dfrac{a(x)}{b(x)}=q(x) + \dfrac{r(x)}{b(x)}$, where the degree of $r(x)$ is less than the degree of $b(x)$. What is the quotient, $q(x)$ ? $ q(x)=$ What is the remainder, $r(x)$ ? $r(x)=$
Explanation: Note that $a(x)$ has a higher degree than $b(x)$. This allows us to find a non-zero quotient polynomial, $q(x)$. [Why is this important?] Let's use long division with polynomials in order to find the quotient, $q(x)$ and remainder, $r(x)$ of $\ \dfrac{a(x)}{b(x)}=\dfrac{3x^4-2x^2+x+5}{x^4+x^2+x+1}$ : We divide ${x^4}$ into ${3x^4}$ to get ${3}$ : $ \hphantom{1567|1444477784} {3}\\ {{{x^4}+x^2+x+1}}|\overline{{3x^4}-2x^2+x+\ 5}\\ \hphantom{37....8888........|}\llap{-}\underline{(3x^4+3x^2+3x+3)}\\ \hphantom{37|3....99888889......}{-5x^2-2x +2}\\ $ [What did we do here?] The process stops here because $x^4+x^2+x+1$ is a polynomial of the fourth degree and $-5x^2-2x +2$ is a polynomial of the second degree. So it follows that ${r(x)}={-5x^2-2x +2}$, ${q(x)}={3}$, and $ \dfrac{3x^4-2x^2+x+5}{x^4+x^2+x+1}={3}+\dfrac{{-5x^2-2x +2}}{x^4+x^2+x+1}$ To conclude, $q(x)=3$ $r(x)=-5x^2-2x +2$